Parametric Equations, Polar Coordinates and Vector-Valued Functions: Unit Diagnostic (Calculus BC Unit 9)

Answer: (B). A vertical tangent needs dx/dt = 0 and dy/dt ≠ 0. dx/dt = 3 − 3t² = 0 at t = ±1, and only t = 1 is allowed. There dy/dt = 4t = 4 ≠ 0. The point is (3 − 1, 2 + 1) = (2, 3).

• (A) uses dy/dt = 0 (t = 0), the condition for a horizontal tangent, and t = 0 is not allowed.
• (C) comes from t = −1, which breaks the condition t > 0.
• (D) swaps the coordinates.

If you missed this: “Horizontal and vertical tangents” in the Topic 9.1 study guide.

Answer: (C). dy/dx = 2t/(3t²) = 2/(3t). Differentiate with respect to t: −2/(3t²). Then divide by dx/dt = 3t²: d²y/dx² = −2/(9t⁴), which is −2/9 at t = 1. Negative, so concave down. (Check: y = x^(2/3), and its second derivative at x = 1 is −2/9.)

• (A) stops after differentiating dy/dx with respect to t and forgets to divide by dx/dt.
• (B) divides y″(t) by x″(t), which is not the method.
• (D) loses the minus sign from differentiating 2/(3t).

If you missed this: “Deriving the correct method” in the Topic 9.2 study guide.

Answer: (D). (dx/dt)² + (dy/dt)² = 4t² + 4t⁴ = 4t²(1 + t²), so the integrand is 2t√(1 + t²) for t ≥ 0. With u = 1 + t², the length is (2/3)(1 + t²)^(3/2) from 0 to √3, which is (2/3)(8 − 1) = 14/3.

• (A) forgets to subtract the value at the lower limit.
• (B) loses the factor 2 when taking the square root of 4t².
• (C) is the straight-line distance from (0, 0) to (3, 2√3). The curve is longer than the chord.

If you missed this: Worked example 1 in the Topic 9.3 study guide.

Answer: (A). By the product rule, x′(t) = e^(−t) − t e^(−t) = (1 − t)e^(−t), so x′(3) = −2e^(−3) < 0: moving left. y′(t) = 2t − 4, so y′(3) = 2 > 0: moving up. The slope is y′/x′ = 2/(−2e^(−3)) = −e³.

• (B) differentiates t e^(−t) as e^(−t), missing the product rule.
• (C) divides x′ by y′.
• (D) misreads the sign of y′(3).

If you missed this: “What r′(t) tells you about the curve” in the Topic 9.4 study guide.

Answer: (C). Use r(2) = r(1) + ∫ from 1 to 2 of v(t) dt, one component at a time. x: 4 + [t²] from 1 to 2 = 4 + 3 = 7. y: −2 + [2 sin(πt/2)] from 1 to 2 = −2 + (0 − 2) = −4.

• (A) treats ⟨4, −2⟩ as the position at t = 0.
• (B) is the displacement from t = 1 to t = 2, not the position.
• (D) drops the chain-rule factor: the antiderivative of π cos(πt/2) is 2 sin(πt/2), not π sin(πt/2).

If you missed this: Worked example 1 in the Topic 9.5 study guide.

Answer: (B). v(1) = ⟨3, 2⟩, so the speed is √(9 + 4) = √13. a(t) = ⟨−2t, 2⟩, so a(1) = ⟨−2, 2⟩. The product v · a = 3(−2) + 2(2) = −2 < 0, so the speed is decreasing.

• (A) assumes the speed rises because both velocity components are positive.
• (C) adds the components instead of using Pythagoras.
• (D) is the square of the speed.

If you missed this: “Is the speed increasing or decreasing?” in the Topic 9.6 study guide.

(a) x′(t) = (t − 1)(t − 3), which is negative between its zeros. The particle moves left for 1 < t < 3.

(b) At rest needs both components zero at the same time. x′ = 0 only at t = 1 and t = 3, and there y′ = 1 and −1. y′ = 0 only at t = 2, where x′ = −1. So the particle is never at rest.

(c) x(3) = 1 + ∫ from 0 to 3 of (t² − 4t + 3) dt = 1 + (9 − 18 + 9) = 1. y(3) = 0 + ∫ from 0 to 3 of (2 − t) dt = 6 − 4.5 = 1.5. Position (1, 1.5).

If you missed this: the sign table in Worked example 1 and “Integrals: displacement, position and distance” in the Topic 9.6 study guide.

Answer: (D). Write x = θ cos θ and y = θ sin θ. By the product rule, dx/dθ = cos θ − θ sin θ = −π/2 and dy/dθ = sin θ + θ cos θ = 1 at θ = π/2. So dy/dx = 1/(−π/2) = −2/π.

• (A) is dr/dθ, the rate of change of the distance from the pole, not the slope.
• (B) divides dx/dθ by dy/dθ.
• (C) confuses the direction of the ray with the direction of the curve.

If you missed this: “Three derivatives, three meanings” in the Topic 9.7 study guide.

Answer: (A). r = 0 when tan θ = −1, at θ = −π/4 and θ = 3π/4, and r > 0 between them. So the circle is traced once for −π/4 ≤ θ ≤ 3π/4. Since r² = 4(1 + sin 2θ), Area = ½ ∫ from −π/4 to 3π/4 of 4(1 + sin 2θ) dθ = 2[θ − ½ cos 2θ] from −π/4 to 3π/4 = 2π. (Check: x² + y² = 2x + 2y is a circle of radius √2.)

• (B) integrates from 0 to 2π. The circle is traced twice on that interval.
• (C) forgets to square r.
• (D) integrates from 0 to π/2 only, which misses part of the circle.

If you missed this: “Choosing the limits: trace the region once” in the Topic 9.8 study guide.

Answer: (B). The curves meet where 5 sin θ = 2 + sin θ, so sin θ = ½ and θ = π/6 or 5π/6. At θ = π/2 the circle (r = 5) is outside the limaçon (r = 3). Area = ½ ∫ from π/6 to 5π/6 of [25 sin²θ − (2 + sin θ)²] dθ = ½ ∫ from π/6 to 5π/6 of (8 − 12 cos 2θ − 4 sin θ) dθ = ½(16π/3 + 6√3 − 4√3) = 8π/3 + √3.

• (A) uses only π/6 to π/2, which is half the region.
• (C) leaves out the ½.
• (D) squares the difference of the radii, (R − r)², instead of using R² − r².

If you missed this: “From one curve to two” in the Topic 9.9 study guide.

(a) 2 − 2 cos θ = 3 gives cos θ = −½, so θ = −2π/3 and θ = 2π/3, the points (−3/2, ±3√3/2).

(b) Test θ = 0: the circle has r = 3 and the cardioid has r = 0. So the circle is outer for −2π/3 < θ < 2π/3. (For other θ the cardioid has r > 3, so R has no points there.)

(c) Area = ½ ∫ from −2π/3 to 2π/3 of [9 − (2 − 2 cos θ)²] dθ = ½ ∫ from −2π/3 to 2π/3 of (3 + 8 cos θ − 2 cos 2θ) dθ = ½(4π + 8√3 + √3) = 2π + 9√3/2 (about 14.08).

If you missed this: Steps 1–3 and Worked example 1 in the Topic 9.9 study guide.